Let CPkk+n=CPk+n/CPk−1. If the bottom-cell inclusion S2k=CPkk→CPkk+2 admits aretraction, then 24 divides k. Writing x=[γ]−1, a retracted Bott generator has the form xk+axk+1+bxk+2. Comparing it with its image under ψ2(x)=2x+x2 gives a=−k/2 and b=k(3k+5)/24; integrality forces both 3 and 8 to divide k.